When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 7.2g Ag and 8.0g

Question

Chemistry

Answer 1

16Ag(s) + S8(s) ==> 8Ag2S(s)  Balanced equation

 

moles Ag present = 7.2 g x 1 mol/108 g = 0.067 mole Ag present

moles S8 present = 8.0 g S8 x 1 mol/256 g = 0.031 moles S8 present

Limiting reactant = Ag

moles Ag2S possible = 0.067 moles Ag x 8 moles Ag2S/16 moles Ag = 0.0335 moles Ag2S

mass Ag2S = 0.0335 moles x 248 g/mol = 8.3 grams 

 

PS.  I've had several interruptions during this, so check my math.

Answer 2

Always write the balanced equation.

2Ag_(s) + S_(s) ---> Ag₂S_(s)

 

We see the molar ratio of Ag:S is 2:1.

 

Calculate the number of moles of each reactant.

Moles of S = 8 g / 32.065 g/mol = 0.249 mol S

Moles of Ag = 7.2 g / 107.87 g/mol = 0.0667 mol Ag

Your limiting reactant is Ag (because you would need 0.249 x 2 or 0.498 mol Ag, which you don't have, to react with sulfur, using the molar ratio).

Now, 2 mol Ag forms 1 mol of Ag₂S.

So, 0.0667 mol Ag forms (0.0667 / 2) mol Ag₂S = 0.03337 mol

Multiplying that by the molar mass of Ag₂S, we get: 0.3337 mol x 247.8 g/mol = 8.269 g Ag₂S